A math 超難ｑｕｅｓｔｉｏｎ

1. It is true for n = 1 since 3/2(3^1-1) = 3
Let it be true for n=r, i.e. 3+3^2+3^3+...+3^r = 3/2(3^r-1)
When n = r+1
3+3^2+3^3+...+3^r + 3^(r+1)
= 3/2(3^r-1) + 3^(r+1)
=1/2(3^(r+1)-3/2 + 3^(r+1)
=3/2(3^(r+1) -3/2
=3/2(3^(r+1)-1)
There fore it is true for n=r+1

Since it is true for n=1, n=r and n=r+1, therefore it is true for all values of n.


2. It is true for n=1 since 2-(1+2/2^1) = (1/2)
Let it be true for n = r, i.e. (1/2)+ (2/2^2)+(3/2^3)+ +(r/2^r) = 2-(r+2/2^r)
when n = r+1
(1/2)+ (2/2^2)+(3/2^3)+ +(r/2^r)+((r+1)/2^(r+1))
= 2-(r+2/2^r) )+((r+1)/2^(r+1))
= 2 -((2r+4-r-1)/2^(r+1))
= 2 -((r+3)/2^(r+1))
= 2 - (((r+1)+2)/2^(r+1))
Therefore it is true for n = r+1

Since it is True for n=1, n= r and n=r+1, therefore it is true for all values of n.

1.3+3^2+3^3+......+3^n=3/2(3^n-1)

Assume the statement is true for n = m, i.e.,
3+3^2+3^3+......+3^m=3/2(3^m-1)

Adding 3^(m + 1) (which is clearly the left-hand side's next term) to both sides does not change the equality:
3+3^2+3^3+......+3^m+3^(m + 1)
=3/2(3^m-1)+3^(m + 1)
=(1/2)3^(m+1)-3/2+3^(m + 1)
=(3/2)3^(m+1)-3/2
=(3/2)[3^(m+1)-1]

We get 3+3^2+3^3+......+3^m+3^(m + 1) =(3/2)[3^(m+1)-1]
Thus, the statement is also true for n = m+1

We can say that 3+3^2+3^3+......+3^n=3/2(3^n-1) is true for all positive integers n


2.(1/2)+(2/2^2)+(3/2^3)+....+(n/2^n)=2 - (n+2)/2^n (This is the correct term for RHS)

Assume the statement is true for n = m, i.e.,
(1/2)+(2/2^2)+(3/2^3)+....+(m/2^m)=2-(m+2)/2^m

Adding (m+1)/2^(m + 1) (which is clearly the left-hand side's next term) to both sides does not change the equality:
(1/2)+(2/2^2)+(3/2^3)+....+(m/2^m) +(m+1)/2^(m + 1)
=2-(m+2)/2^m+(m+1)/2^(m + 1)
=2 - 2(m+2)/2(2^m) + (m+1)/2^(m + 1)
=2 - (2m+4)/2^(m+1) + (m+1)/2^(m + 1)
=2 + [m+1 - (2m+4)]/2^(m + 1)
=2 + [m+1 - 2m-4)]/2^(m + 1)
=2 + [-m-3]/2^(m + 1)
=2 - [m+3]/2^(m + 1)
=2 - [(m+1) + 2]/2^(m + 1)

We get (1/2)+(2/2^2)+(3/2^3)+....+(m/2^m) +(m+1)/2^(m + 1)=2 - [(m+1) + 2]/2^(m + 1)Thus, the statement is also true for n = m+1

We can say that (1/2)+(2/2^2)+(3/2^3)+....+(n/2^n)=2 - (n+2)/2^n is true for all positive integers n

1.
Let P(n)be the proposition "3+3^2+3^3+...+3^n=3/2(3^n-1)" for all positive integers n
When n = 1, LHS = 3^1=3
RHS = 3/2(3^1-1)=3
So P(1) is true.
Assume P(k) is true for any positive integer k
i.e. 3+3^2+3^3+...+3^k=3/2(3^k-1)
when n= k+1 ,
3+3^2+3^3+...+3^k+3^(k+1)
= 3/2(3^k-1)+3^(k+1)
= 3/2[3^k-1+2*3^k]
= 3/2[3*3^k -1]
= 3/2[3^(k+1) -1]
P(k+1)is true if P(k)is true for any positive integer k
So, by MI, P(n)is true for all positive integer n.

2.
Let P(n) be the proposition "(1/2)+(2/2^2)+(3/2^3)+....+(n/2^n)=2-[(n+2)/2^n]"
When n=1, LHS = 1/2
RHS = 2-[(1+2)/2^1]=2-3/2=1/2
So P(1) is true
Assume P(k) is true for any positive integer k
i.e. (1/2)+(2/2^2)+(3/2^3)+....+(k/2^k)=2-[(k+2)/2^k]
when n=k+1,
(1/2)+(2/2^2)+(3/2^3)+....+(k/2^k)+[(k+1)/2^(k+1)]
= 2-[(n+2)/2^n]+[(k+1)/2^(k+1)]
= 2- [ (2k+4-k-1)/2^(k+1)]
= 2- [(k+3)/2^(k+1)]
P(k+1)is true if P(k)is true for any positive integer k
So, by MI, P(n)is true for all positive integer n.